Take a look at this post : https://www.beatthegmat.com/integers-pro ... tml#320994Ramit88 wrote:Does the integer g have a factor f such that 1<f<g ?
1. g>3!
2. 11! + 11 >= g >=11! + 2
The problems are almost same.
factorial
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Rahul@gurome
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The problems are almost same.
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aleph777
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Want to be sure I have a grasp on this solution. Can someone confirm?
The question is essentially asking: is G prime?
STATEMENT 1 in insufficient because: g>3! means g could be ANY number greater than 1 * 2 * 3. Therefore, g could be 8, 9, or 10, in which case it would have a smaller factor that is greater than 1. But it could also be 7, 11, etc., in which case f would not exist.
STATEMENT 2 is sufficient, however, because: g could be one of 10 different consecutive integers that are composed of 11! plus some number between 2 and 11, which means the number being added can never push the new value into a prime since it's appearing both in the factorial and as an addition, and is therefore a multiple.
The question is essentially asking: is G prime?
STATEMENT 1 in insufficient because: g>3! means g could be ANY number greater than 1 * 2 * 3. Therefore, g could be 8, 9, or 10, in which case it would have a smaller factor that is greater than 1. But it could also be 7, 11, etc., in which case f would not exist.
STATEMENT 2 is sufficient, however, because: g could be one of 10 different consecutive integers that are composed of 11! plus some number between 2 and 11, which means the number being added can never push the new value into a prime since it's appearing both in the factorial and as an addition, and is therefore a multiple.
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Exactly, exactly right. Kudos!aleph777 wrote:Want to be sure I have a grasp on this solution. Can someone confirm?
The question is essentially asking: is G prime?
STATEMENT 1 in insufficient because: g>3! means g could be ANY number greater than 1 * 2 * 3. Therefore, g could be 8, 9, or 10, in which case it would have a smaller factor that is greater than 1. But it could also be 7, 11, etc., in which case f would not exist.
STATEMENT 2 is sufficient, however, because: g could be one of 10 different consecutive integers that are composed of 11! plus some number between 2 and 11, which means the number being added can never push the new value into a prime since it's appearing both in the factorial and as an addition, and is therefore a multiple.
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alextheodosi
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Hi,
I have just done this problem. I agree with statement 1.
Regarding statement 2:
when i look at this problem
I see that 11! is common to both sides of this inequality, therefore we can subtract 11! from both sides can't we?
And then 11>=g>=2? which presents the same as statement 1, ie includes some primes and if G can be prime than the statement is not sufficient.
While i understand the explanation below, I jumped to seeing the above (is this mathematically correct? I think it is? and if it is then why does it change the outcome of the problem?)
I don't understand why the above is not correct?
Thanks,
Alex
I have just done this problem. I agree with statement 1.
Regarding statement 2:
when i look at this problem
I see that 11! is common to both sides of this inequality, therefore we can subtract 11! from both sides can't we?
And then 11>=g>=2? which presents the same as statement 1, ie includes some primes and if G can be prime than the statement is not sufficient.
While i understand the explanation below, I jumped to seeing the above (is this mathematically correct? I think it is? and if it is then why does it change the outcome of the problem?)
I don't understand why the above is not correct?
Thanks,
Alex
